chapter 5 3 4 first and second fundamantal theorem微积分第一基本定理

5.3TheFirstFundamentalTheoremof a,b]，then bexistsanditsvariablewouldbedenotedast，suchas ft"x˛a,

subintervala,x, xthevalue

ft

correspondstothevalueofx

x=

ftisanewfunctiondefinedbyfona,bandnamedtheAccumulationfunctionofforUpper-limitfunctionoff.yf

yox1x fyox1xxx=x

ftx12=0tdt=2x12

˛TheoremAFirstFundamentalTheoremofLetlet

becontinuousontheclosedinterval[a,b]andbea(variable)pointin(a,b).Then axax

f(t)dt=f(x)Thatmeanstheaccumulationfunctionantiderivativeoff(x)

f

isExample

(1)y=xet2 y=1ln

,find

= =

=-ln=-

dx Solution:BytheFirstFundamentalTheoremofdxt3dt

Example

ddx

242t.nx

ucosuduSolution:NotethefactthatxisthelowerlimitandusetheFundamentalTheoremofd4tan

ucosudu=d-xtan2ucosudu

dx =-dxtan2ucosudu

ProofoftheFirstFundamentalTheoremofx

LetG(x)=

f(t)dtxG(x+h)-G(x)=

f(t)dt-

f

f(t)dt

f(t)dt-

f(t)dt

f Letf(1)= f(t),f(t2)= f t˛[x, t˛[1)h (1)h f(t)£G(x+h)-G(x)£f(t (1)2)=hfi Example

t u BytheFirstFundamentalTheoremofCalculus,notethattheupperlimitisx2 ratherthanx.uLetu=

,F(u)

t theChainrule,weF[u(x)]=(u)u=u3t =(3u-1)2x=2x(3x2-CorollaryoftheFundamentalTheoremofff˛C[a,b],gandharebothdifferentiablex,andg(x),h(x)˛[a,b]¢h(fg(x)g(x)h(x)(x). Letcbeafixedpointin

h(

f(t)dt= f(t)dt+h(

f(t)dt= f(t)dt-

fuLetF(u)=u

f(t)dt,andu=g(x),UsingtheChainruleandtheFirstFundamentalTheoremofCalculus,wegetcc

f(t)dt

Fg(x)=(u)g(x)(x)=[g(x)]g(x=cc

x)f

¢=f[h(x)]h(x).¢

g(x)fh(

g(xg(x)

h(x(x).ExampleLetf(x)=

tan2

(t2-1)dt.

(x)x (x)x)-t=(tan2x-1)sec2x-2x(x4-txExamplex

xfi

SolutionThisistheindeterminateform 00xUsingL’Hopital’sRule,wextxtx

xsin x x

xfi

xfi

xfi0+x=1limx

=11=1xx xfi xxSymmertyofdefiniteintgral f˛Ca,a,yOaxIffsanyOaxaa-a xdx=a0fIffxsanoddfunction, aa fx= 1p1

Example

xcosxdx (a)Becausexcosxisanoddfunctionon-p,pp 2

xcosxdx=

2(b)Becauseln2+xisanoddfunctionon1,1,2-x1

dx=(f(x)=ln2+x,f(-x)=ln2-x=-ln2+x=-f(x),2-x 2+x 2-xln2+xisanoddfunction.）2-x5.4TheSecondFundamentalTheoremofCalculusandtheMethodofSubstitutionTheoremASecondFundamentalTheoremof let

becontinuous(henceintegrable)on[a,b],andbeanyantiderivativeoff on[a,b].Thenb f(x)dx=F(b)-FbbThisformulab

f(x)dx=F(b)-Fa.aisalsocalledTheNewton-Labniz'sFormulaandisa.abexpressedasb

f(x)dx=F(x)b

orF(x) Recall5.3TheoremA(TheFirstTheoremofCalculus),weknowxF(x)= f(t)dtisanantiderivativeoff(x)i.eF(x)=fForanyF(x),antiderivativeoff(x),wehaveF(x)=Letx=a

F(x)-F(x)”F(x)=F(x)+F(a)=F(a)+andnotethatF(a)= C=-F F(x)=F(x)+C=F(x)-F(a),thatx f(t)dt=F(x)-F

letx=b,theproofisBecauseF(x)isanantiderivativeoff(x),i.e.F(x)=f cingf(x)withF(x),wecanrewritetheb(x)dxF(x)dx,badF(x)=F(b)-FbTheImportance N-L•Provideapowerfultoolforevaluating•MakethelinkbetweendifferentiationandExample

Show

bkdx=k(b- ,wherea

isSolution:F(x)=kxisanantiderivativeof f(x)=k.Thus,bytheSecondFundamentalTheoremofCalculus,bkdx=x

=kb-ka=k(b-Exampleandr„-1

bxrdx,a

risareal

F(x)

xrr

isanantiderivative

f

Thus,bytheSecondFundamentalTheoremofb xr+1 ar br+1-arb xdx= r r r rExample

Evaluate24x-6x2

24x-6x2dx=2x2-2x3 82 2 24x-6x2dx=42xdx-6x2dx2 x2 x3 1 =42 -63 --6 +

2

4318(431Example

1

x3+ .8x

+x43dx=3x43

+3x738 8=316+3128-31+3 =45+381=6519

1e2x

1 10 +x0111e2x

dx=1e2x+ln(1+0

1+x 0 =1e2+ln2-1+0

-12

EvaluatingDefiniteTheoremTheoremASubstitutionRuleforindefiniteLetfbeadifferentiablefunctionandsupposethatFisanantiderivativeoff.Then,ifu=g(x),g(x)g(x)dxg(x)+Proof.Itisenoughtoshowthatthederivativeoftherightmemberis theintegrandoftheleftmember.Butthatisasimpleapplicationof theChainRulecombinedwiththefactthatF¢=f g(x)g(x)g(x)g(x)

x x2x2+ 2.x2+x2+

,thendu

dx.x2+x2x2+x2+x2+

dx=sinudu=-cosu+x2+=-x2+4Example 4

x2+x. Step Letu=x2+x,thendu=2x

.{ 1422x1{ 142u

udu=2u32+C=2x2+x32+ 44 2x+1dx2344 =22032-0= p0Example sin3 0 Letu=sin ,thendu=2cos2xdx.sin32

(sin2x)32cos2x 14 14=1u3du14

+C= 2x+p

2 sin2xx Question:CanwecombinetheabovetwostepsintoYes!ThisiswhatthefollowingThmTheoremTheoremBSubstitutionRuleforDefiniteLetghaveacontinuousderivativeon,andletbecontinuousontherangeofba g(x)g(x)dx.fLetFisanantiderivative f,therightsideofg(ag(a

f(u)du=[F(u)]g(b)=F(g(b))-F(g(a)).AndbecauseF(g(x))isanantiderivativeof f(g(x))g(xsotheleftsideofequation f(g(x))g(x)d=[F(g(x))]a=F(g(b))-F(g(a)).x2+4Werecallthex2+4Example

2.1dxLetu=x2+x,thendu=2x1dx x=0«u=0,x=4«u=44

x2+(2xx2+(14243

udu

2u32 5=232-0=5 1Example 1

Letu=x2+2x+6,2x2dxx1dxnote

x=0«u= andx=1«u=9.

dx=

2x+1)20x2

+2x+6

2

x

+2x+191= 1du=- 26u2 6=-11-1=1 Example

p2 44 xx9 Letu=x,thenx=u2,dx x «u=,x «u p2 4 dx=2 2uduxup2 xu 2cos3p2sinpp

=2-g(x)g(x)dx示成f(u)du后再积分.这就要求解题者熟练掌握基本初等xex2

dx.222 22x

dx=ex eudu

u=12

u

+C=1ex2+2⑵ x2lnx+1)

=2lnx+1 2lnx 2lnx

1u

u=2ln=1ln2

u=2ln

2

2lnx+1+tanxdxcot

sinx

dx=-

u=cosx-1u=-⑵

+C=

cotxdx=cosxdx= sin sin=lnsinx+

utantanxdx=lnsecx+cotxdx=-lncscx+例求积分⑴

a⑵

aa2-a2-a2+

dx=a2x

xa1+ a

dx

a x a

a1+1+ a=1arctanx+C. 1a2+dx=1arctanx+aaa2-1-axaa2-1-axa1-x2a1-x2aa

1-u21-u2

=arcsinx+a dx=arcsinx+Ca2-a例求积分 dxa>

dx

x2

xx=

-

x+a=1

dx

dx dx=1 dx=1lnx-a+x2-

+C=1

例secxdx,cscsecxdx

cosxdx=

dsin

cos2

1-sin2

du=

+ du=1 1-1+1+sin1-sin

1+1+1-+1-sin2

1+u

+C= 21+sinx cosx2

secx+tanx+secxdx=lnsecx+tanx+⑵

dx

sinxdx=-

sin

sin2

1-cos2

=1 =12

u

2-cos-cos1-cos22=1

1-cosx sin

+C=

cscx-cotx+ cscxdx=lncscx-cotx+tanxdtanxdx=lnsecx+=-lncscx+1a2+dx=1arctanx+ a2-aaxa x2-secdx1=lnsecx+tanx+=lncscx-cotx+xx

x1+

xdxx

xx xx x+

1+(xxux

2arctanudu=2arctanud(arctanu